Let a, b, c and d be positive real numbers such that a + b + c + d...

Given:

$a + b + c + d = 11$

We know that,

$AM \ge GM$

$⇒ \dfrac{\dfrac{a}{5} + \dfrac{a}{5} + \dfrac{a}{5} + \dfrac{a}{5} + \dfrac{a}{5} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{c}{2} + \dfrac{c}{2} + \dfrac{d}{1}}{11} \ge \dfrac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}11$

$⇒ \dfrac{a + b + c + d}{11} \ge \dfrac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}11$

$⇒ \dfrac{11}{11} \ge \dfrac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}11$

$⇒ \dfrac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}11 \le 1$

$⇒ {a}^{5}{b}^{3}{c}^{2}d \le {5}^{5}{3}^{3}{2}^{2}$

$⇒ {a}^{5}{b}^{3}{c}^{2}d \le 90 \cdot 3750$

So,

$\beta = 90$